# Solving a cubic system of equations

I have the following equation system:

$A_1 x + B_1 y + C_1 z + D_1 xy + E_1 xz + F_1 yz + G_1 xyz = M_1$ $A_2 x + B_2 y + C_2 z + D_2 xy + E_2 xz + F_2 yz + G_2 xyz = M_2$ $A_3 x + B_3 y + C_3 z + D_3 xy + E_3 xz + F_3 yz + G_3 xyz = M_3$

$A_1$, $B_1$, ..., $M_1$, $A_2$, $B_2$, ..., $M_2$, $A_3$, $B_3$, ..., $M_3$ are known.

Trying to get $x$ based on $y$ and $z$ from the first equation, then substituting it in the second equation, then getting $y$ based on $z$ and substituting it in the third equasion seems a nightmare. How to solve this equation system?

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Where did you find this question? If I may ask. – 1015 Feb 16 '13 at 14:32
I think we need to know more about the coefficients. As far as I know, there isn't a general solution to this type of equation system. – Flavin Feb 16 '13 at 14:47
It's part of another problem I'm working on. See my another question: math.stackexchange.com/questions/305395/… – Tamás Pap Feb 16 '13 at 15:02
@Flavin: The problem is that to solve my actual problem I have to be able to solve it without knowing the coefficients. Actually this is related to another problem: math.stackexchange.com/questions/305395/… – Tamás Pap Feb 16 '13 at 15:04
@vonbrand: Big thanks for the edit! – Tamás Pap Feb 16 '13 at 15:22

A possible workaround can be the following:

1. Let's multiply the first equation with with $-G_2$, the second equation with $G_1$, and add them.
2. Now multiply the second equation with $-G_3$, the third one with $G_2$, and add them.
3. Finally multiply the third equation with $-G_1$, and the first one with $G_3$ and add them.

Theoretically we have now tree new equations, but we escaped from the $xyz$ part.

In the same way let's escape from xy, xz, xz.

We have now a simple equation system with 3 equations and 3 unknowns:

$AA_1x + BB_1y + CC_1z = MM_1$

$AA_2x + BB_2y + CC_2z = MM_2$

$AA_3x + BB_3y + CC_3z = MM_3$

that's easy to solve.

If you think this wouldn't always work please let me know, or if you have a better solution it is welcomed!

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It will not work. Try to solve with your method the system of equations $x+y+xy=5$ and $x-y+2xy=3$. (Or even more tiny example for meditation: $x+x^{17}=4$.) – zaarcis Feb 16 '13 at 19:43
You are probably right. Can you give an equation system with 3 unknowns similar to system in the question, where this will not work? Thanks! – Tamás Pap Feb 16 '13 at 20:23
If lots of $0$-s as coefficients is okay, then $x+xyz=7$, $y+xyz=8$, $z+xyz=9$. You can't make intended transformations simultaneously, and making them one by one leaves you (before last transformation) with only one equation with $xyz$ - and you can't make $1\cdot k_1 + 0\cdot k_2$ equal to $0$ (if $k_1\neq 0$ and $k_2\neq 0$, of course). It means you can't get rid of all $xyz$-s by such method. – zaarcis Feb 16 '13 at 20:31
Yeah, you are right! Any ideas? – Tamás Pap Feb 16 '13 at 20:58
Use numerical methods to approximate solution. (And no, I can't tell more because I'm weak at them.) – zaarcis Feb 17 '13 at 14:30

Maybe a bit ambiguous, but you can also try to set up the equations for $M_1+M_2$, $M_1+M_3$, $M_2+M_3$ and $M_1+M_2+M_3$, which is easy, in total you then have 7 equations, with 7 unknowns ($x,y,z,xy,xz,yz,xyz$). But i dont think this is even allowed.

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It's allowed but it will help nothing. It will just make 4 more equations with no new information. – zaarcis Feb 17 '13 at 14:34